To solve the problem, we need to find the sum of the intercepts on the coordinate axes of the plane that passes through the point \((-2, -2, 2)\) and contains the line joining the points \((1, -1, 2)\) and \((1, 1, 1)\).
### Step-by-Step Solution:
1. **Identify the Points**:
- Point \( O = (-2, -2, 2) \)
- Point \( P = (1, -1, 2) \)
- Point \( Q = (1, 1, 1) \)
2. **Find Direction Vectors**:
- The vector \( \overrightarrow{OP} = P - O = (1 - (-2), -1 - (-2), 2 - 2) = (3, 1, 0) \)
- The vector \( \overrightarrow{OQ} = Q - O = (1 - (-2), 1 - (-2), 1 - 2) = (3, 3, -1) \)
3. **Calculate the Normal Vector**:
- The normal vector \( \mathbf{n} \) of the plane can be found using the cross product \( \overrightarrow{OP} \times \overrightarrow{OQ} \):
\[
\mathbf{n} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & 1 & 0 \\
3 & 3 & -1
\end{vmatrix}
\]
- Expanding this determinant:
\[
\mathbf{n} = \mathbf{i} \left(1 \cdot (-1) - 0 \cdot 3\right) - \mathbf{j} \left(3 \cdot (-1) - 0 \cdot 3\right) + \mathbf{k} \left(3 \cdot 3 - 1 \cdot 3\right)
\]
\[
= -\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}
\]
- Thus, \( \mathbf{n} = (-1, 3, 6) \).
4. **Equation of the Plane**:
- The equation of the plane can be written as:
\[
-1(x + 2) + 3(y + 2) + 6(z - 2) = 0
\]
- Simplifying this:
\[
-x - 2 + 3y + 6 + 6z - 12 = 0 \implies -x + 3y + 6z - 8 = 0
\]
- Rearranging gives:
\[
x - 3y - 6z = -8
\]
5. **Finding Intercepts**:
- The intercepts on the axes can be found from the equation:
\[
\frac{x}{-8} + \frac{y}{\frac{8}{3}} + \frac{z}{\frac{8}{6}} = 1
\]
- Thus, the intercepts are:
- \( x \) intercept: \( -8 \)
- \( y \) intercept: \( \frac{8}{3} \)
- \( z \) intercept: \( \frac{8}{6} = \frac{4}{3} \)
6. **Sum of the Intercepts**:
- The sum of the intercepts is:
\[
-8 + \frac{8}{3} + \frac{4}{3} = -8 + \frac{12}{3} = -8 + 4 = -4
\]
### Final Answer:
The sum of the intercepts on the coordinate axes of the plane is \(-4\).
